Answer
The pH of this solution is equal to $4.87$.
Work Step by Step
1. Find the numbers of moles:
- $50ml = 0.05L$
$C(CH_3COOH) * V(CH_3COOH) = 0.15* 0.05 = 7.5 \times 10^{-3}$ moles
$C(CH_3COO^-) * V(CH_3COO^-) = 0.2* 0.05 = 0.01$ moles
2. Calculate the total volume:
- Total volume: 0.05 + 0.05 = 0.1L
3. Calculate the final concentrations:
$[CH_3COOH] : \frac{7.5 \times 10^{-3}}{0.1} = 0.075M$
$[CH_3COOH] : \frac{0.01}{0.1} = 0.1M$
4. Calculate the pKa Value
$pKa = -log(Ka)$
$pKa = -log( 1.8 \times 10^{- 5})$
$pKa = 4.745$
5. Check if the ratio is between 0.1 and 10:
- $\frac{[Base]}{[Acid]} = \frac{0.1}{0.075}$
- 1.333: It is.
6. Check if the compounds exceed the $K_a$ by 100 times or more:
- $ \frac{0.1}{1.8 \times 10^{-5}} = 5556$
- $ \frac{0.075}{1.8 \times 10^{-5}} = 4167$
7. Using the Henderson–Hasselbalch equation:
$pH = pKa + log(\frac{[Base]}{[Acid]})$
$pH = 4.745 + log(\frac{0.1}{0.075})$
$pH = 4.745 + 0.1249$
$pH = 4.87$
